SME
Physical Chemistry I: Thermodynamics & Kinetics · Problem type
Unimolecular rate against pressure (Lindemann)
Use this for “For A + M ⇌ A′ + M then A′ → P, k₁ = 10⁶ M⁻¹s⁻¹, k₋₁ = 10⁹ M⁻¹s⁻¹ and k₂ = 10⁶ s⁻¹. Find kᵤₙᵢ at [M] = 10⁻³ M, and the high-pressure limit.”
Chart
kᵤₙᵢ(1) = 500
This is an example. Type your own values to start a new problem.
k₁
Rate constant of energizing
example · M⁻¹s⁻¹
k₋₁
Rate constant of calming
example · M⁻¹s⁻¹
k₂
Rate constant of reacting
example · s⁻¹
[M]
Bath gas concentration
example · M
kᵤₙᵢ
Unimolecular rate constant
calculated · s⁻¹
k∞
High-pressure limit
calculated · s⁻¹
[M]½
Bath gas for half of k∞
worked out (not typed) · M
Formulas
kᵤₙᵢ
=
k₁
×
k₂
×
[M]
÷
(k₋₁
×
[M]
+
k₂
)
500
=
1 ×
10
6
×
(1
×
10
6
)
×
0.001
1 ×
10
9
× 0.001 + (1 ×
10
6
)
k∞
=
k₁
×
k₂
k
₋₁
1,000
=
1 ×
10
6
×
1 ×
10
6
1 ×
10
9
[M]
1
2
=
k₂
k
₋₁
0.001
=
1 ×
10
6
1 ×
10
9
Assumptions
•
Collisions with the bath gas M energize A to A′ (k₁) and calm it again (k₋₁); the energized A′ alone falls apart to products (k₂).
•
A′ is at steady state: it is made as fast as it is used, so the rate is kᵤₙᵢ[A] with kᵤₙᵢ = k₁k₂[M] ÷ (k₋₁[M] + k₂).
•
At high pressure kᵤₙᵢ levels off at k∞ = k₁k₂ ÷ k₋₁ (first order); at low pressure it is k₁[M] (second order).
Step-by-step
WE KNOW
k₁ = 1 × 10⁶ M⁻¹s⁻¹, k₋₁ = 1 × 10⁹ M⁻¹s⁻¹, k₂ = 1 × 10⁶ s⁻¹, [M] = 0.001 M
FIND
unimolecular rate constant (kᵤₙᵢ), high-pressure limit (k∞), bath gas for half of k∞ ([M]½)
Step 1 · Find unimolecular rate constant (kᵤₙᵢ)
Use kᵤₙᵢ = k₁ × k₂ × [M] ÷ (k₋₁ × [M] + k₂)
A′
is
made
at
k₁
[M]
and
shared
between
calming,
k₋₁[M],
and
reacting,
k₂
:
the
share
that
reacts
times
k₁
[M].
kᵤₙᵢ
=
10
6
×
10
6
×
0.001
1 ×
10
9
× 0.001 +
10
6
kᵤₙᵢ
=
10
6
×
10
6
×
0.001
1,000,000 +
10
6
kᵤₙᵢ
=
10
6
×
10
6
×
0.001
2,000,000
kᵤₙᵢ
=
10
12
×
0.001
2,000,000
kᵤₙᵢ
=
10
9
2,000,000
kᵤₙᵢ = 500 s⁻¹
Step 2 · Find high-pressure limit (k∞)
Use k∞ = k₁ × k₂ ÷ k₋₁
At
high
[M],
k₋₁[M]
swamps
k₂
and
the
[M]
cancels:
the
limit
k₁
k₂
k
₋₁.
k∞
=
10
6
×
10
6
10
9
k∞
=
10
12
10
9
k∞ = 1,000 s⁻¹
Step 3 · Find bath gas for half of k∞ ([M]½)
Use [M]½ = k₂ ÷ k₋₁
Where
calming,
k₋₁[M],
is
as
fast
as
reacting,
k₂
,
half
of
A′
reacts:
kᵤₙᵢ
is
half
of
k∞.
[M]
1
2
=
10
6
10
9
[M]
1
2
=
0.001
M
Check
Put the numbers back in. Both sides should match.
500
=
10
6
×
10
6
×
0.001
1 ×
10
9
× 0.001 +
10
6
✓
1,000
=
10
6
×
10
6
10
9
✓
0.001
=
10
6
10
9
✓
Topic check
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